Hyperstatic

A structure is known as hyperstatic if and only if the number of independent connections which it dregs is higher than the number of its degrees of freedom.

The degree of hyperstaticity of a structure is the numbers of connections to remove to obtain a structure Isostatique.

Calculation of the hyperstatic elastic structures

The calculation of the hyperstatic elastic structures is carried out classically by considering the beam with a cut which in limit the number of connection and makes the beam hyperstatic. This imaginary structure causes reactions of supports which one can calculate with the equilibrium equations static. One finds also the constraints and the deformation within this imaginary structure.

The real structure is not this imaginary structure: additional connections binds the elements of them. One calculation then efforts of supports necessary to respect the connections. This rectification is done at the price of a certain effort in the beam and of deformations which one can calculate.

In virtues of the principle of superposition, these constraints and these deformations must be added with that with the imaginary hyperstatic structure, and the constraints and real structural deformation are found.

Embed-supported beam requested in its center

In the case of a beam embed-supported solicited in its center of an effort $P$ (fig. 1), we calculate a beam first of all Bi-supported (fig.2).

We find the reactions of supports equal each one to half of the effort $P$ and an angular deformation in $O$ and the swing angle in $O$,

$\backslash \; phi\_0\; =\; \backslash \; frac\; \{P\; L^2\}\; \{the\; 16th\; I\}$

Remain to calculate the moment of embedding at the point $A$. This moment is the required effort to reach an angle of null deformation in $A$, i.e. the Bi-supported moment causing a deformation opposed exactly to that caused by the load on a beam which we have just calculated. If we believe the traditional formulas of them, this moment is worth

$M\_O\; =\; \backslash \; frac\; \{3\; \backslash \; phi\_0\; E\; I\}\; \{L\}\; =\; \backslash \; frac\; \{3\; P\; L\}\; \{16\}$

Thus, we calculated all the reactions of supports and can calculate the efforts and the resulting deformations in the beam like summons effort in the isostatic beam and that of with the additional reaction of supports which we introduced.

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